[Tutor] Pi approximation

[boB Stepp]

I see I wrote the below a little too quickly!  Don't forget to take
the reciprocal when printing.  You might want to modify my naming of
variables to reflect this.  And return the reciprocal, which actually
gives the pi approximation in the function form.

On Wed, Mar 28, 2018 at 9:08 PM, boB Stepp <robertvstepp at gmail.com> wrote:
> On Wed, Mar 28, 2018 at 2:09 PM, Roger Lea Scherer <rls4jc at gmail.com> wrote:
>> In one of my lessons I am asked to compare approximations for pi. I got
>> everything to work properly and my attempt is successful and matches
>> Python's approximation up to 15 digits to the right of the decimal, but I
>> suspect I can do this programmatically rather than the repetitious way I
>> did.
>>
>> I tried "for i in range(10):"; then I tried "c += c" so it would be a sum.
>> Those attempts did not work. I tried math.fsum and though the documentation
>> says it is for iterables I could not get it to work as I desired. I
>> received an error that said TypeError: 'float' object is not iterable
>>
>> I included all the code so I wouldn't neglect any you might need. Can you
>> help again?
>
> You are using the formula in the Wikipedia article, right?  Mimic in
> Python what you are doing by manually.  Do something like:
>
> pi_approx = 0.0
> for k in range(10):
>     pi_approx += your formula from the Wikipedia article
>
> print(pi_approx)
>
> You might even want to make a function out of the above so that you
> can try iterating over different ending values for k:
>
> def calc_pi(loop_value):
>     pi_approx = 0.0
>     for k in range(loop_value):
>         pi_approx += your formula
>     return pi_approx
>
> print(calc_pi(10))
>
> You actually had all the pieces mentioned.  You just need to put them
> together, looping just like you would do if you were calculating by
> hand.
>
> HTH!
>
> boB
>
>>
>> # compare various approximations of pi
>> import math
>> import random
>>
>> # simplest estimate
>> a = 22/7
>> print(a)
>>
>> # next simplest
>> b = 355/113
>> print(b)
>>
>> # from wikipedia:
>> # In 1910, the Indian mathematician Srinivasa Ramanujan found several
>> rapidly converging infinite series
>> c = (2*math.sqrt(2)/9801) * (((math.factorial(4*0))*(1103+26390*0)) /
>> ((math.factorial(0)**4)*(396**(4*0))))
>> d = (2*math.sqrt(2)/9801) * (((math.factorial(4*1))*(1103+26390*1)) /
>> ((math.factorial(1)**4)*(396**(4*1))))
>> e = (2*math.sqrt(2)/9801) * (((math.factorial(4*2))*(1103+26390*2)) /
>> ((math.factorial(2)**4)*(396**(4*2))))
>> f = (2*math.sqrt(2)/9801) * (((math.factorial(4*3))*(1103+26390*3)) /
>> ((math.factorial(3)**4)*(396**(4*3))))
>> g = (2*math.sqrt(2)/9801) * (((math.factorial(4*4))*(1103+26390*4)) /
>> ((math.factorial(4)**4)*(396**(4*4))))
>> h = (2*math.sqrt(2)/9801) * (((math.factorial(4*5))*(1103+26390*5)) /
>> ((math.factorial(5)**4)*(396**(4*5))))
>> i = (2*math.sqrt(2)/9801) * (((math.factorial(4*6))*(1103+26390*6)) /
>> ((math.factorial(6)**4)*(396**(4*6))))
>> j = (2*math.sqrt(2)/9801) * (((math.factorial(4*7))*(1103+26390*7)) /
>> ((math.factorial(7)**4)*(396**(4*7))))
>> k = (2*math.sqrt(2)/9801) * (((math.factorial(4*8))*(1103+26390*8)) /
>> ((math.factorial(8)**4)*(396**(4*8))))
>> l = (2*math.sqrt(2)/9801) * (((math.factorial(4*9))*(1103+26390*9)) /
>> ((math.factorial(9)**4)*(396**(4*9))))
>> m = c + d + e + f + g + h + i + j + k + l
>> print(1/m)
>>
>> print(math.pi)



-- 
boB